MATH 2107 LINEAR ALGEBRA II ASSIGNMENT 1 DUE: October 16 at the beginning of the tutorial 1. Let 32:PPT→be a linear transformation s.t.32)(xxT=, 0)1(=+xT, xxT=−)1(.Find Tand determine)1(2++xxT. The set }1,1,{2−+xxxis a basis of2P, so every vector 2cxbxap++=in 2Pis a linear combination of these vectors. i.e. )1()1()(11212−+++=++=xcxbxacxbxapfor some scalar 1a,1b,1c. i.e. 2111112)()(xaxcbcbcxbxa+++−=++i.e. we need to solve ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−+⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−2/)(2/)(100010001~001110110abbaccba( several steps of row operations need to be taken to reach the final RREF of the matrix). i.e. )1(2)()1(2)()(22−−++++=++=xabxbaxccxbxapSo, )1(2)()1(2)()()(2−−++++=xTabxTbaxcTpTxabcxxabbacxcxbxaT2)(2)(02)()(332−+=−+++=++⇒So, 3322)11(.1)1(xxxxxT=−+=++2. Show that the function ),(),(xyyxyxT+=is not a linear transformation. I. Let ),(111yxv=and ),(222yxv=So,),()(11111yxyxvT+=, ),()(22222yxyxvT+=),(212121yyxxvv++=+And )))((,()(2121212121yyxxyyxxvvT+++++=+),()()(2211212121yxyxyyxxvTvT++++=+And, )()()(2121vTvTvvT+≠+. So the given function is not closed under vector addition. II. Let ),(yxv=. So ),(ayaxav=),()(2xyaayaxavT+=⇒),()(xyyxvT+=),()(axyayaxvaT+=⇒So the given function is not closed under scalar multiplication. Hence the given function is not a linear transformation. (By showing that any oneof the operations is not satisfied, you can claim that it is not a linear transformation)

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3. Let 34:ℜ→ℜTbe a function given by )7542,542,432(),,,(wzyxwzyxwzyxwzyxT+++++−−+++=a) Show that Tis a linear transformation. Let ),,(11,111wzyxv=and ),,(22,222wzyxv=be two vectors in 4ℜ, I. Then ),,,(2121212121wwzzyyxxvv++++=+),(4)(3)(2)(()(2121212121wwzzyyxxvvT+++++++=+),(5)(4)(2)(21212121wwzzyyxx+++++−+−))(7)(5)(4)(221212121wwzzyyxx+++++++)7542,542,432()(1111111111111wzyxwzyxwzyxvT+++++−−+++=)7542,542,432()(2222222

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